Question

if $f(x)=cos x - 6\tan x$, then $f(x)=$ $f(2)=$

Explanation:

Step1: Recall derivative rules

Use the sum - difference rule of derivatives $(u - v)'=u' - v'$, and the derivative formulas $(\cos x)'=-\sin x$ and $(\tan x)'=\sec^{2}x$.

Step2: Differentiate $f(x)$

If $f(x)=\cos x - 6\tan x$, then $f'(x)=(\cos x)' - 6(\tan x)'$. Substituting the derivative formulas, we get $f'(x)=-\sin x-6\sec^{2}x$.

Step3: Evaluate $f'(2)$

Substitute $x = 2$ into $f'(x)$. First, recall that $\sec x=\frac{1}{\cos x}$, so $f'(2)=-\sin(2)-6\sec^{2}(2)=-\sin(2)-\frac{6}{\cos^{2}(2)}$. Using a calculator (assuming $x$ is in radians), $\sin(2)\approx0.9093$ and $\cos(2)\approx - 0.4161$, then $\cos^{2}(2)\approx0.1731$, and $f'(2)=- 0.9093-\frac{6}{0.1731}\approx-0.9093 - 34.662\approx - 35.571$.

Answer:

$f'(x)=-\sin x - 6\sec^{2}x$
$f'(2)\approx - 35.571$